根据我上次问的问题:Applying PageRank to a topic hierarchy tree(using SPARQL query extracted from DBpedia)
因为我目前得到了针对Regulated concept map的PageRank值。关于“机器学习”的概念,我目前的代码如下:
from SPARQLWrapper import SPARQLWrapper, N3
from rdflib import Graph, URIRef, Literal
import networkx as nx
from networkx.readwrite import json_graph
from rdflib.extras.external_graph_libs import rdflib_to_networkx_graph
from rdflib.namespace import Namespace, RDFS, FOAF
import matplotlib.pyplot as plt
#SPARQL query for Regulated SPARQL Query Strategy
sparql = SPARQLWrapper("http://dbpedia.org/sparql")
sparql.setQuery("""construct { ?child skos:broader <http://dbpedia.org/resource/Category:Machine_learning> . ?gchild skos:broader ?child }
where {
{ ?child skos:broader <http://dbpedia.org/resource/Category:Machine_learning> . ?gchild skos:broader ?child}
UNION
{ ?gchild skos:broader/skos:broader <http://dbpedia.org/resource/Category:Machine_learning> . ?gchild skos:broader ?child}
}
""")
sparql.setReturnFormat(N3)
results = sparql.query().convert()
g = Graph()
g.parse(data=results, format="n3")
#Undirected graphs will be converted to a directed graph with two directed edges for each undirected edge.
dg = rdflib_to_networkx_graph(g, False, edge_attrs=lambda s,p,o:{})
#Draw regulated concept map
nx.draw(dg)
plt.draw()
#PageRank calculation
p1 = nx.pagerank(dg, alpha=0.85)
#p1 to pr(dict to list)
pr = sorted(p1.items(), key=lambda x:x[1],reverse=True)[:10]
#print sorted ranking
for key,val in pr:
print(key,val)
有几个问题:
{ ?child skos:broader <http://dbpedia.org/resource/Category:Machine_learning> . ?gchild skos:broader ?child}
分配为绿色,而{ ?gchild skos:broader/skos:broader <http://dbpedia.org/resource/Category:Machine_learning> . ?gchild skos:broader ?child}
分配为红色李>#Draw regulated concept map
# nx.draw(dg,pos=nx.spring_layout(dg),node_color='red') # use spring layout
# edges = nx.draw_networkx_edges(dg,pos=nx.spring_layout(dg))
pos = nx.spring_layout(dg)
source_node=copy.copy(pos)
print(source_node)
source_node_list = list(source_node.keys())
# print(source_node_list[0] in nx.spring_layout(dg))
# print(source_node_list)
options = {"node_size": 25, "alpha": 0.85}
graph=nx.draw_networkx_edges(dg, pos=pos, width=1.0, alpha=0.5)
graph=nx.draw_networkx_nodes(dg, pos=pos, nodelist=[source_node_list[1]], node_color="r", **options)
graph=nx.draw_networkx_nodes(dg, pos=pos, nodelist=[source_node_list[0],], node_color="b", **options)
graph=nx.draw_networkx_nodes(dg, pos=pos, nodelist=source_node_list[2:len(source_node_list)-1], node_color="g", **options)
# nx.draw(graph)
# plt.draw()
# nx.draw_networkx_edges(
# dg,
# pos,
# edgelist=[source_node_list[1]],
# width=8,
# alpha=0.5,
# edge_color="r",
# )
# nx.draw_networkx_edges(
# dg,
# pos,
# edgelist=[source_node_list[0]],
# width=8,
# alpha=0.5,
# edge_color="b"
# )
# nx.draw_networkx(dg, pos=nx.spring_layout(dg), node_color='blue',with_labels = False)
# labels=nx.draw_networkx_labels(dg,pos=nx.spring_layout(dg))
# nodes = nx.draw_networkx_nodes(dg,pos=nx.spring_layout(dg))
# nx.draw(dg)
# plt.draw()
事先非常感谢
我认为可以将字典传递给draw函数的
node_color
参数。如果构造该字典时,键是节点名,值是要与这些节点名关联的颜色,则应该能够获得所需的格式例如,如果您已经能够运行一些SPARQL来生成一个您希望为绿色的节点列表和另一个您希望为蓝色的节点列表,并且假设您已经有了这些节点名称的
green_list
和blue_list
对列表,那么您可以这样构造dict:理想情况下,您可以在绘图时将node_colors_d放入您的参数中,它应该为您着色。从内存来看,一些nx版本更喜欢颜色为delivered as a list,其顺序与节点名称相同——但我认为这应该适用于当前版本
或者,假设nx需要一个列表,那么您可以用
node_colours_l
替换node_colours_d
来执行相同的作业,方法是替换以下内容:这将创建一个列表,其中包含按顺序映射到图形中每个节点外观的颜色,并且您将此列表提交给draw函数的node_color参数
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